Math Problem Statement

IQ is normally distributed with a mean of 100 and a standard deviation of 15. Use the Cumulative Z-Score Table to answer the questions below.Hint

Suppose one individual is randomly chosen. Find the probability that this person has an IQ greater than 119.5. Write your answer in percent form with two decimal places.

P(IQ greater than 119.5) = %

Suppose one individual is randomly chosen. Find the probability that this person has an IQ less than 95.5. Write your answer in percent form with two decimal places.

P(IQ less than 95.5) = %

In a sample of 800 people, how many people would have an IQ less than 140.5? Round your answer to the nearest whole number.

people

In a sample of 800 people, how many people would have an IQ greater than 100? Round your answer to the nearest whole number.

people

Solution

We will calculate the required probabilities and values using the properties of the normal distribution and the cumulative Z-Score Table.

1. Probability of IQ greater than 119.5

To calculate this:

  • Mean (μ\mu) = 100
  • Standard Deviation (σ\sigma) = 15
  • Value of interest (XX) = 119.5

The Z-score is calculated as: Z=XμσZ = \frac{X - \mu}{\sigma} Z=119.510015=19.5151.30Z = \frac{119.5 - 100}{15} = \frac{19.5}{15} \approx 1.30

Using a Z-Score Table:

  • The cumulative probability for Z=1.30Z = 1.30 is approximately 0.90320.9032.
  • The probability for P(Z>1.30)P(Z > 1.30) is: P(Z>1.30)=10.9032=0.0968P(Z > 1.30) = 1 - 0.9032 = 0.0968

Convert to percent: P(IQ>119.5)=9.68%P(IQ > 119.5) = 9.68\%


2. Probability of IQ less than 95.5

Here:

  • X=95.5X = 95.5

Calculate the Z-score: Z=Xμσ=95.510015=4.5150.30Z = \frac{X - \mu}{\sigma} = \frac{95.5 - 100}{15} = \frac{-4.5}{15} \approx -0.30

Using a Z-Score Table:

  • The cumulative probability for Z=0.30Z = -0.30 is approximately 0.38210.3821.

Convert to percent: P(IQ<95.5)=38.21%P(IQ < 95.5) = 38.21\%


3. Number of people with IQ less than 140.5 in a sample of 800

Here:

  • X=140.5X = 140.5

Calculate the Z-score: Z=Xμσ=140.510015=40.5152.70Z = \frac{X - \mu}{\sigma} = \frac{140.5 - 100}{15} = \frac{40.5}{15} \approx 2.70

Using a Z-Score Table:

  • The cumulative probability for Z=2.70Z = 2.70 is approximately 0.99650.9965.

The number of people in the sample: Number=0.9965×800797\text{Number} = 0.9965 \times 800 \approx 797


4. Number of people with IQ greater than 100 in a sample of 800

Here:

  • The mean is 100100, so Z=0Z = 0.
  • The cumulative probability for Z=0Z = 0 is 0.50000.5000.

The probability for P(Z>0)P(Z > 0) is: P(Z>0)=10.5000=0.5000P(Z > 0) = 1 - 0.5000 = 0.5000

The number of people in the sample: Number=0.5000×800=400\text{Number} = 0.5000 \times 800 = 400


Final Answers

  1. P(IQ>119.5)=9.68%P(IQ > 119.5) = 9.68\%
  2. P(IQ<95.5)=38.21%P(IQ < 95.5) = 38.21\%
  3. Number of people with IQ less than 140.5: 797797
  4. Number of people with IQ greater than 100: 400400

Do you want detailed explanations or need help with similar calculations?

Here are 5 related questions to explore:

  1. What is the probability of having an IQ between 90 and 110?
  2. How would the probabilities change if the standard deviation were 10 instead of 15?
  3. What is the Z-score for an IQ of 130, and how do you interpret it?
  4. In a sample of 1,000 people, how many would you expect to have an IQ between 85 and 115?
  5. What percentile rank corresponds to an IQ of 105?

Tip: Always double-check the Z-Score Table to ensure accurate cumulative probabilities.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-Score Calculation
Probability
Sampling in Statistics

Formulas

Z = (X - μ) / σ
P(Z > z) = 1 - P(Z ≤ z)
Number = Probability × Sample Size

Theorems

Properties of the Normal Distribution
Empirical Rule for Normal Distributions

Suitable Grade Level

Grades 10-12